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\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle RDK rectangle en R, on sait que :
\begin{itemize}
\item RD=7,3 cm
\item \(\widehat{DKR}\) =  16\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [KD]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle DZL rectangle en D, on sait que :
\begin{itemize}
\item DL=4,3 cm
\item ZL=9,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DLZ}\).

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle VGR rectangle en V, on sait que :
\begin{itemize}
\item GR=6,4 cm
\item \(\widehat{GRV}\) =  41\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [VR]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle FMJ rectangle en F, on sait que :
\begin{itemize}
\item MJ=5,5 cm
\item \(\widehat{FMJ}\) = 59\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [FM]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle SGA rectangle en S, on sait que :
\begin{itemize}
\item SG=3,3 cm
\item GA=8,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{SGA}\). \newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

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 Dans le triangle RDK rectangle en R, on cherche une relation entre l'angle aigu   \(\widehat{RKD}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{RKD}\)) = \(\dfrac{RD}{DK}\)
\par\addvspace{.5em}
 D'où sin(16\textdegree{}) = \(\dfrac{7{,}3}{DK}\)
\par\addvspace{.5em}
On a donc DK =\(\dfrac{7{,}3}{\sin(16^{\circ})}\) \(\approx\) 26,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle DZL rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DLZ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{DLZ}\)) = \(\dfrac{DL}{ZL}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{DLZ}\)) =  \(\dfrac{4{,}3}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{DLZ}\) = Arccos (\(\dfrac{4{,}3}{9{,}5}\)) \(\approx\) 63\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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 Dans le triangle VGR rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VRG}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{VRG}\)) = \(\dfrac{VR}{GR}\)
\par\addvspace{.5em}
  d'où    cos(41\textdegree{}) =  \(\dfrac{VR}{6{,}4}\)

On a donc VR = 6,4 \(\times\) cos(41\textdegree{}) \(\approx\) 4{,}8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle FMJ rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FMJ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{FMJ}\)) = \(\dfrac{FM}{MJ}\)
\par\addvspace{.5em}
d'où cos(59\textdegree{}) = \(\dfrac{FM}{5{,}5}\)

On a donc FM = 5,5 \(\times\) cos(59\textdegree{}) \(\approx\) 2,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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 Dans le triangle SGA rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SGA}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{SGA}\)) = \(\dfrac{SG}{GA}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SGA}\)) = \(\dfrac{3{,}3}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SGA}\) =ArcCos(\(\dfrac{3{,}3}{8{,}7}\)) \(\approx\)68\textdegree{}\end{document}
